Question

已知f（x）=mx/x²+n（m,n∈R）在x=1处取得极值2.求f（x）的解析式?

Answer 1

f'(x)=(nx-mx^2)/(x^2+n)^2 f'(1)=(n-m)/(1+n)^2=0 m=n

f(1)=m/(1+n)=m/(1+m)=2 m=-2 n=-2

f(x)=-2x/(x^2-2)